Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. Concerning the Fibonacci sequence, prove that \(F_{1}+F_{3}+F_{5}+F_{7}+\cdots+F_{2 n-1}=F_{2 n}\).

Short answer

The given statement \(F_{1}+F_{3}+F_{5}+F_{7}+...+F_{2n-1} = F_{2n}\) regarding the Fibonacci sequence is proven to be true through mathematical induction. The base case for \(n = 1\) is established and using the inductive hypothesis and the property of the Fibonacci sequence, the statement is proven to be true for \(n+1\) assuming it's true for \(n\).

Step-by-step solution

Understanding Fibonacci Sequence

The Fibonacci sequence is a series of numbers where every number after the first two is the sum of the two preceding ones. It usually starts with 0 and 1. Here, the sequence starts with \(F_{1}, F_{2}, F_{3},... F_{n}\) where each \(F_{i}\) represents the ith number in the Fibonacci sequence and \(F_{i} = F_{i-1} + F_{i-2}\). The problem is focusing on the odd indexed Fibonacci numbers and their sums.

Base Case

In proof by induction, before the general case, we need to show that the statement holds for the smallest possible value. This is known as the base case. Here, we let \(n = 1\). We look at the left hand side (LHS) of the equation \(F_{1} = 1\) and the right hand side (RHS) \(F_{2} = 1\) and find out they are equal which verifies that the statement is true for \(n = 1\). Hence, the base case is established.

Inductive Step

The inductive step is where we make an assumption and then we try to prove the statement for \(n+1\) assuming it's true for \(n\). Assume that the statement holds for some positive integer \(n\), which means \(F_{1}+F_{3}+F_{5}+...+F_{2n-1} = F_{2n}\). We need to prove that the statement holds for \(n+1\), i.e. \(F_{1}+F_{3}+...+ F_{2n-1} + F_{2n+1} = F_{2n+2}\). We know that \(F_{2n+2} = F_{2n+1} + F_{2n}\). Replacing \(F_{2n}\) with \(F_{1}+F_{3}+F_{5}+...+F_{2n-1}\) (by inductive assumption) in the equation, we get \(F_{2n+2} = F_{2n+1} + F_{1}+F_{3}+F_{5}+...+ F_{2n-1}\), which is similar to the original equation, hence proven.